Question:
The electric fields of two plane electromagnetic plane waves in vacuum are given by
\(\vec{E_1}=E_0\hat{j}\cos(\omega t-kx)\) and
\(\vec{E_2}=E_0\hat{k}\cos(\omega t-ky)\)
At \(t=0\), a particle of charge \(q\) is at origin with a velocity \(\vec{v}=0.8c\,\hat{j}\) (\(c\) is the speed of light in vacuum). The instantaneous force experienced by the particle is :
The electric fields of two plane electromagnetic plane waves in vacuum are given by
\(\vec{E_1}=E_0\hat{j}\cos(\omega t-kx)\) and
\(\vec{E_2}=E_0\hat{k}\cos(\omega t-ky)\)
At \(t=0\), a particle of charge \(q\) is at origin with a velocity \(\vec{v}=0.8c\,\hat{j}\) (\(c\) is the speed of light in vacuum). The instantaneous force experienced by the particle is :
\(\vec{E_1}=E_0\hat{j}\cos(\omega t-kx)\) and
\(\vec{E_2}=E_0\hat{k}\cos(\omega t-ky)\)
At \(t=0\), a particle of charge \(q\) is at origin with a velocity \(\vec{v}=0.8c\,\hat{j}\) (\(c\) is the speed of light in vacuum). The instantaneous force experienced by the particle is :
Show Hint
Find the magnetic field of each wave from \(\vec B=\frac{1}{c}\hat{n}\times\vec E\) and use \(\vec F=q(\vec E+\vec v\times\vec B)\) at \(t=0\) at the origin.
Updated On: Nov 5, 2026
- \(E_0q\left(0.8\hat{i}-\hat{j}+0.4\hat{k}\right)\)
- \(E_0q\left(-0.8\hat{i}+\hat{j}+\hat{k}\right)\)
- \(E_0q\left(0.8\hat{i}+\hat{j}+0.2\hat{k}\right)\)
- \(E_0q\left(0.4\hat{i}-3\hat{j}+0.8\hat{k}\right)\)
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The Correct Option is C
Solution and Explanation
Wave 1 travels along +x with \(\vec E_1=E_0\hat{j}\): \(\vec B_1=\frac{1}{c}\hat{i}\times E_0\hat{j}=\frac{E_0}{c}\hat{k}\).
Wave 2 travels along +y with \(\vec E_2=E_0\hat{k}\): \(\vec B_2=\frac{1}{c}\hat{j}\times E_0\hat{k}=\frac{E_0}{c}\hat{i}\).
At the origin, \(t=0\): \(\vec E=E_0(\hat{j}+\hat{k})\), \(\vec B=\frac{E_0}{c}(\hat{i}+\hat{k})\).
\(\vec v\times\vec B=0.8c\,\hat{j}\times\frac{E_0}{c}(\hat{i}+\hat{k})=0.8E_0(-\hat{k}+\hat{i})\).
\(\vec F=qE_0\left[\hat{j}+\hat{k}+0.8\hat{i}-0.8\hat{k}\right]=E_0q\left(0.8\hat{i}+\hat{j}+0.2\hat{k}\right)\).
Answer: (3) \(E_0q(0.8\hat{i}+\hat{j}+0.2\hat{k})\)
Wave 2 travels along +y with \(\vec E_2=E_0\hat{k}\): \(\vec B_2=\frac{1}{c}\hat{j}\times E_0\hat{k}=\frac{E_0}{c}\hat{i}\).
At the origin, \(t=0\): \(\vec E=E_0(\hat{j}+\hat{k})\), \(\vec B=\frac{E_0}{c}(\hat{i}+\hat{k})\).
\(\vec v\times\vec B=0.8c\,\hat{j}\times\frac{E_0}{c}(\hat{i}+\hat{k})=0.8E_0(-\hat{k}+\hat{i})\).
\(\vec F=qE_0\left[\hat{j}+\hat{k}+0.8\hat{i}-0.8\hat{k}\right]=E_0q\left(0.8\hat{i}+\hat{j}+0.2\hat{k}\right)\).
Answer: (3) \(E_0q(0.8\hat{i}+\hat{j}+0.2\hat{k})\)
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